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Ok so I mean greater than either of the other 2 not the combination. I created a grid and found that there are 15/36 possibilities that it would be greater than one of the dice, so was thinking that it would be 30/72 that it would be greater than one or the other?

Elite Member

Ok so I mean greater than either of the other 2 not the combination. I created a grid and found that there are 15/36 possibilities that it would be greater than one of the dice, so was thinking that it would be 30/72 that it would be greater than one or the other?

I still do not understand the question. Is one die special (maybe red and the other two are green) and you want to know if you roll all three dice the probability that the red die is smaller than each green die? Can we please see your work showing us how you got 15/36? Please!?

Elite Member

Ok so I mean greater than either of the other 2 not the combination. I created a grid and found that there are 15/36 possibilities that it would be greater than one of the dice, so was thinking that it would be 30/72 that it would be greater than one or the other?

Thanks for the clarification. Rolling three dice given \(\displaystyle 6^3=216\) outcomes in the form of triples.
Now must consider the numbers, look at this list:
\(\displaystyle \begin{array}{*{20}{c}}{\text{max value}}&{\text{#}}&{\text{total}}\\\hline{6|}&{25}&{75}\\{5|}&{16}&{48}\\{4|}&9&{27}\\{3|}&4&{12}\\{2|}&1&3\end{array}\)
The table tells us that there are seventy-five triples in which six it the greatest value.
That that is: the triple \(\displaystyle (6,1,5) \) has a die with a value greater than either of the other two.
BUT that happens in three ways. Look at the set of pairs \(\displaystyle \{1,2,3,4,5\}\times\{1,2,3,4,5\}\) there are twenty-five pairs.
Each of those pairs can go with the six in three ways. (Can you tell why we din't to use the one?)
Note that we counted the triple \(\displaystyle (4,5,4)\) but not \(\displaystyle (4,4,1)\) WHY?
Add up the third column. What do you get?
That is, we count the triples having a an unique maximum.
What is the answer to this problem?

My previous post was written before you completely clarified the setup.
However, the model is the same. The outcomes are still triples. But now the question is to count the triples in which the third entry is the unique maximum. This table tells us all:
\(\displaystyle \begin{array}{*{20}{c}}{\text{max value}}&{\text{# of triples}}\\\hline{6|}&{25}\\{5|}&{16}\\{4|}&9\\{3|}&4\\{2|}&1\end{array}\)
Now the sum of the second column is \(\displaystyle 55\).
That means that in rolling a die three times there are \(\displaystyle 55\) times that the third value is the unique maximum of the three.
So if we roll two red dice and then roll one white die, what is the probability that the value on the white die is greater than the values on either of the red dice?